Question

a certain mass of gas occupies 300 cm3 at 50°C at what temperature will it have its volume reduce by half? assuming the pressure remain constant

Answer 1

Volume 1 = V1 = 300 cm^3

Temperature 1 = T1 = 50°C

Volume 2 = V2 = V1/2 (given)

Temperature 2 = T2 = ?

by using Charles' law, where the pressure remains constant (∆P = 0), we get

V/T = constant

or, V1/T1 = V2/T2

by putting the respected values, we get

(300)/(50) = (300/2)/T2

or, 6 = 150/T2

or, T2 = 150/6

or, T2 = 25

At 25° C, the volume of the gas reduces by half.

Answer 2

Answer:

The temperature of gas when its volume reduced to half is equal to -111.5°C.

Explanation:

Given, the initial volume of gas, V₁ = 300cm³

The temperature of the gas, T₁ = 50°C = 273 + 50 = 323K

The final volume of the gas, V₂ = 300/2 = 150cm³

Consider that, T₂ is the final temperature of the gas.

According to Charles's law, at constant pressure, the volume of the given mass of the gas is directly proportional to absolute temperature.

V ∝ T

$\frac{V_1}{T_1} =\frac{V_2}{T_2}$

$T_2=\frac{V_2\times T_1}{V_1}$

$T_2=\frac{(150)(323)}{(300)}$

$T_2=161.5K$

As we know, T(°C) = T (kelvin) - 273

$T_2 = 161.5 -273 =-111.5^oC$

Therefore, the the final temperature of gas is -111.5°C or 161.5K.

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